Q 49. - Calculus

Question

In Exercises $43-50$, compute all of the second-order partial derivatives for the functions $f\left(r,\theta \right)=r\sin \theta$ and show that the mixed partial derivatives are equal. 

Step-by-Step Solution

Answer

The second order partial derivatives for the function are

$$\begin{gathered} \frac{{\partial }^{2}f}{\partial r^2}=0 \\ \frac{{\partial }^{2}f}{\partial {\theta }^2}=-r\sin \theta \end{gathered}$$

Step 1. Definition

Clairaut's theorem on equality of mixed partials states that under assumption of continuity of both the second-order mixed partials of a function of two variables, the two mixed partials are equal. 

Step 2. Finding second order partial derivative

$f\left(r,\theta \right)=r\sin \theta \cdots \cdots \left(1\right)$
Partially differentiate equation $\left(1\right)$ both sides with respect to $r$

$\Rightarrow \frac{\partial f}{\partial r}=\sin \theta$

Again partially differentiate both sides with respect to $r$

$\Rightarrow \frac{{\partial }^{2}f}{\partial r^2}=0$

Partially differentiate equation $\left(1\right)$ both sides with respect to $\theta$

$\Rightarrow \frac{\partial f}{\partial \theta }=r\cos \theta$

Again partially differentiate both sides with respect to $\theta$

$\Rightarrow \frac{{\partial }^{2}f}{\partial {\theta }^2}=-r\sin \theta$

Step 3. Finding mixed order partial derivative

$f\left(r,\theta \right)=r\sin \theta$

$$\begin{gathered} \Rightarrow \frac{{\partial }^{2}f}{\partial r\partial \theta }=\frac{\partial }{\partial r}\left(r\cos \theta \right) \\ \Rightarrow \frac{{\partial }^{2}f}{\partial r\partial \theta }=\cos \theta \end{gathered}$$

Also

$$\begin{gathered} \Rightarrow \frac{{\partial }^{2}f}{\partial \theta \partial r}=\frac{\partial }{\partial \theta }\left(\sin \theta \right) \\ \Rightarrow \frac{{\partial }^{2}f}{\partial \theta \partial r}=\cos \theta \end{gathered}$$

Now as we observe

$\frac{{\partial }^{2}f}{\partial r\partial \theta }=\frac{{\partial }^{2}f}{\partial \theta \partial r}$ [Hence proved]